Answer:
V2 = 90.7 mL
Explanation:
pressure and volume are inversely proportional, if the pressure is increased, the volume will decrease. In an isothermal process:
p1V1 = p2V2
V2 = p1V1/p2 = (277 torr×187 mL)/571 torr
V2 = 90.7 mL
Answer:
liquid oxygen is highly flammable
Explanation:
near any source of heat. liquid oxygen can explode to flames thus being hazardous
<h3><u>Answer;</u></h3>
<em>All the above</em>
Workers at construction sites often reduce erosion by;
- <em>Moving excess sediment back to its original location
</em>
- <em>Planting trees
</em>
- <em>Spraying water on bare soil</em>
<h3><u>Explanation;</u></h3>
- Soil erosion is a naturally occurring process which involves the wearing away of the topsoil by natural forces such as wind, water or other forces associated with farming.
- <em><u>Construction of roads and buildings results to large amounts of soil erosion around the world. It is therefore important to put measures that would help reduce soil erosion at construction sites</u></em>. These measures uses principals of soil control such as implementing sediment control, limiting soil exposure, reducing the runoff velocity, and modifying topography among others.
Answer:
3 M
Explanation:
Molarity equation: M = n/v
n = moles of solute
v = liters of solution
9 moles of NaCl / 3 L
9/3 = 3 M
Answer:
Molecular formula
Explanation:
Molecular formula in the first place is required to understand which compound we have. We then should refer to the periodic table and find the molecular weight for each atom. Adding individual molecular weights together would yield the molar mass of a compound.
Then, dividing the total molar mass of a specific atom by the molar mass of a compound and converting into percentage will provide us with the percentage of that specific atom.
E. g., calculate the percent composition of water:
- molecular formula is
; - calculate its molar mass: [tex]M = 2M_H + M_O = 2\cdot 1.00784 g/mol + 16.00 g/mol = 18.016 g/mol;
- find the percentage of hydrogen: [tex]\omega_H = \frac{2\cdot 1.00784 g/mol}{18.016 g/mol}\cdot 100 \% = 11.19 %;
- find the percentage of oxygen: [tex]\omega_O = \frac{16.00 g/mol}{18.016 g/mol}\cdot 100 \% = 88.81 %.
